Areas between a curve and the y-axis: integrating with respect to y, the sign cases, and using the reflection of y = a^x and y = log_a x in y = x (new in the 2024 Mathematics Advanced syllabus)

The curve meets the $y$-axis where $4 - y^2 = 0$, so $y = \pm 2$, and $x \geq 0$ between them. Hence

Markers expect the intercepts $y = \pm 2$ as the limits, the integral with respect to $y$, and the correct evaluation (using symmetry, $2\int_0^2 (4 - y^2)\,dy$, is also fine).

The region lies between $y = 1$ (where the curve meets the $y$-axis) and $y = e$, to the left of the curve. Integrating in $y$ would need a primitive of $\ln y$, so integrate in $x$ instead: for $0 \leq x \leq 1$ the region runs from the curve up to $y = e$.

Markers award marks for identifying the region's bounds, a correct integral (in either variable, with the reflection argument if in $y$), and the exact answer $1$.

The curve is to the right of the $y$-axis ($x = y^2 \geq 0$), so integrate $x$ with respect to $y$:

Marker's note: one mark for the integral in $y$ with the correct limits, one for $9$.

Rewrite with $x$ as the subject. $x = y^{1/3}$.

Marker's note: one mark for $x = y^{1/3}$ and the integral, one for $\frac{45}{4}$ (note $8^{4/3} = 16$).

Where does the curve meet the $y$-axis? $x = 0$ when $y(y - 4) = 0$, so $y = 0$ and $y = 4$.

Which side? For $0 < y < 4$, $x = y(y - 4) < 0$: the region is to the left of the $y$-axis, so the integral is negative and the area is its absolute value.

The area is $\frac{32}{3}$ square units.

Marker's note: one mark for the limits $0$ and $4$, one for the integral, one for taking the positive area.

Rewrite. $y = \ln x$ means $x = e^y$, which is positive, so the region is right of the $y$-axis.

Marker's note: one mark for $x = e^y$, one for the integral with respect to $y$, one for $e^2 - 1$.

Split where $x$ changes sign. $x = y^3 < 0$ for $-1 \leq y < 0$ and $x > 0$ for $0 < y \leq 2$.

Marker's note: one mark for splitting at $y = 0$, one for each part correct. Integrating straight from $-1$ to $2$ gives $\frac{15}{4}$, which is wrong because the left part cancels.

Interpret the integral as an area

$\ln x \geq 0$ for $1 \leq x \leq e$, so $\int_1^e \ln x \, dx$ is the area under $y = \ln x$ from $x = 1$ to $x = e$.

Use the rectangle

The rectangle $0 \leq x \leq e$, $0 \leq y \leq 1$ has area $e$. It splits into the region under $y = \ln x$ (to the right of the curve) and the region between $y = \ln x$ and the $y$-axis for $0 \leq y \leq 1$.

The second region, integrating in $y$

Since $x = e^y$ there, its area is $\int_0^1 e^y \, dy = e - 1$.

Subtract

$\int_1^e \ln x \, dx = e - (e - 1) = 1$.

Marker's note: one mark for the rectangle, one for $\int_0^1 e^y \, dy$, one for evaluating it, one for the final subtraction. A labelled sketch makes this argument much clearer.

With respect to $y$. $y = \sqrt{x}$ gives $x = y^2$ for $y \geq 0$. The region runs from $y = 0$ to $y = 2$:

Check with respect to $x$. The line $y = 2$ meets the curve at $x = 4$. $R$ is the rectangle $0 \leq x \leq 4$, $0 \leq y \leq 2$ minus the area under $y = \sqrt{x}$:

Marker's note: two marks for each method. Both give $\frac{8}{3}$ square units.

The region. $y = \log_2 x$ meets the $x$-axis at $x = 1$ and reaches $y = 3$ at $x = 8$. The required region is under the curve for $1 \leq x \leq 8$.

Rectangle minus the region beside the $y$-axis. The rectangle $0 \leq x \leq 8$, $0 \leq y \leq 3$ has area $24$. The part of it to the left of the curve is bounded by $x = 2^y$, the $y$-axis, $y = 0$ and $y = 3$:

Subtract. The required area is $24 - \frac{7}{\ln 2} \approx 13.90$ square units.

Marker's note: one mark for the rectangle, one for rewriting as $x = 2^y$, one for $\int 2^y \, dy = \frac{2^y}{\ln 2}$, one for the final answer.